Question: $ g(x) = \int_{0}^{x}\sqrt{5+4\tan {t}}\,dt\,$ $ g\,^\prime\Big(\dfrac{\pi}{4}\Big)\, =$
Explanation: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{5+4\tan {t}}$ is continuous on $[0,\dfrac{\pi}{4}]$. Applying the theorem We're given: $ g(x) = \int_{\,1}^{\,x}\sqrt{5+4\tan{t}}\,dt $ So the theorem tells us: $ g\,^\prime(x) =\sqrt{5+4\tan{x}}$ Evaluating $g'\left(\dfrac\pi4\right)$ $ g\,^\prime\Big(\dfrac{\pi}{4}\Big)= \sqrt{5+4\cdot \tan({\pi}/4)}=\sqrt{5+4\cdot 1}=\sqrt{9}=3$ The answer: $g'\left(\dfrac\pi4\right)=3$